good morning. I have a matrix (eout) with 14610 rows and 16 column.
14610 row represent one of each day of the period between 01-01-1960 and 31-12-2000.
What I need is a new matrix with 40 rows and 16 columns with a mean for each year.
something like a mean of 365 rows continuously.
The problem I have are the leap years each 4 years.
Suggestions to solve this?
For a start, to get the number of days for a given year:
function n = ndays(year)
tmp = repmat([1,1,0,0,0],numel(year),1);
n = datenum([year(:)+1,tmp])-datenum([year(:), tmp]);
end
With this, you could collect the rows e.g. with mat2cell:
rows_per_year = ndays(1960:2000);
chunks = mat2cell(yourInputMatrix, rows_per_year, size(yourInputMatrix,2));
means = cellfun(#(x) mean(x,1), chunks);
(The latter part comes untested..)
Related
I have three vectors of the same size, pressure, year and month. Basically I would like to create a matrix of pressure values that correspond with the months and years that they were measured using a for loop. It should be 12x100 in order to appear as 12 months going down and 100 years going left to right.
I am just unsure of how to actually create the matrix, besides creating the initial structure. So far I can only find pressure for a single month (below I did January) for all years.
A = zeros([12, 100]);
for some_years = 1900:2000
press = pressure(year == some_years & month == 1)
end
And I can only print the pressures for January for all years, but I would like to store all pressures for all months of the years in a matrix. If anyone can help it would be greatly appreciated. Thank You.
Starting with variables pressure, year, and month. I would do something like:
A fairly robust solution using for loops:
T = length(pressure); % get number of time periods. I will assume vectors same length
if(length(pressure) ~= T || length(month) ~= T)
error('length mismatch');
end
min_year = min(year); % this year will correspond to index 1
max_year = max(year);
A = NaN(max_year - min_year + 1, 12); % I like to initialize to NaN (not a number)
% this way missing values are NaN
for i=1:T
year_index = year(i) - min_year + 1;
month_index = month(i); % Im assuming months run from 1 to 12
A(year_index, month_index) = pressure(i);
end
If you're data is SUPER nicely formatted....
If your data has NO missing, duplicate, or out of order year month pairs (i.e. data is formatted like):
year month pressure
1900 1 ...
1900 2 ...
... ... ...
1900 12 ...
1901 1 ...
... ... ...
Then you could do the ONE liner:
A = reshape(pressure, 12, max(year) - min(year) + 1)';
Hi so I am new to MATLAB. I am trying to find the means of weight values for each month over five years and put these values into a matrix that will be 5x12 in size.
I am attempting to accomplish this with a loop but I'm having a little trouble, if anyone can push me towards the right direction that would be awesome, thanks. What I have so far is this:
weight_data = (10 weights per month for 10 years, 1200 weights total)
year = (years 2000-2010) %year 1-10 corresponds with the 1200 weights)
month = (months 1-12) %weights for all months (120 months, correspond with 1200 weights)
weight_vec = zeros([12, 5]);
for n = year(1:5)
weights = weight_data(n);
mean_weights = mean(weights);
end
This only gives me one number though, I assume the mean from the 5 years I'm trying to loop through. I also know I need to incorporate the months somehow but I'm just confused on how to do this.
Being your weight_data matrix (10 weights x 12 months x 10 years), the weight of the i-th year are located in weight_data(:,:,i).
w=weight_data(:,:,i)
w is a (10 x 12) which contains the 10 weights values of the 12 months.
You can use mean to compute the mean value of the weights of each month:
w=mean(weight_data(:,:,i))
Therefore you can setup a loop over the years:
for i=1:1:5
mean_weights(:,i)=mean(weight_data(:,:,i))'
end
(the ' in mean(weight_data(:,:,i))' is required to transpose the output of mean from a row-array to a column-array so that it fits in your output matrix which is (12 x 5)
Hope this helps,
Qapla'
I have observed daily data that I need to compare to generated Monthly data so I need to get a mean of each month over the thirty year period.
My observed data set is currently in 365x31 with rows being each day (no leap years!) and the extra column being the month number (1-12).
the problem I am having is that I can only seem to get a script to get the mean of all years. ie. I cannot figure how to get the script to do it for each column separately. Example of the data is below:
1 12 14
1 -15 10
2 13 3
2 2 37
...all the way to 12 for 365 rows
SO: to recap, I need to get the mean of [12; -15; 13; 2] then [14; 10; 3; 37] and so on.
I have been trying to use the unique() function to loop through which works for getting the number rows to average but incorrect means. Now I need it to do each month(28-31 rows) and column individually. Result should be a 12x30 matrix. I feel like I am missing something SIMPLE. Code:
u = unique(m); %get unique values of m (months) ie) 1-12
for i=1:length(u)
month(i) = mean(obatm(u(i), (2:31)); % the average for each month of each year
end
Appreciate any ideas! Thanks!
You can simply filter the rows for each month and then apply mean, like so:
month = zeros(12, size(obatm, 2));
for k = 1:12
month(k, :) = mean(obatm(obatm(:, 1) == k, :));
end
EDIT:
If you want something fancy, you can also do this:
cols = size(obatm, 2) - 1;
subs = bsxfun(#plus, obatm(:, 1), (0:12:12 * (cols - 1)));
vals = obatm(:, 2:end);
month = reshape(accumarray(subs(:), vals(:), [12 * cols, 1], #mean), 12, cols)
Look, Ma, no loops!
I have 19 cells (19x1) with temperature data for an entire year where the first 18 cells represent 20 days (each) and the last cell represents 5 days, hence (18*20)+5 = 365days.
In each cell there should be 7200 measurements (apart from cell 19) where each measurement is taken every 4 minutes thus 360 measurements per day (360*20 = 7200).
The time vector for the measurements is only expressed as day number i.e. 1,2,3...and so on (thus no decimal day),
which is therefore displayed as 360 x 1's... and so on.
As the sensor failed during some days, some of the cells contain less than 7200 measurements, where one in
particular only contains 858 rows, which looks similar to the following example:
a=rand(858,3);
a(1:281,1)=1;
a(281:327,1)=2;
a(327:328,1)=5;
a(329:330,1)=9;
a(331:498,1)=19;
a(499:858,1)=20;
Where column 1 = day, column 2 and 3 are the data.
By knowing that each day number should be repeated 360 times is there a method for including an additional
amount of every value from 1:20 in order to make up the 360. For example, the first column requires
79 x 1's, 46 x 2's, 360 x 3's... and so on; where the final array should therefore have 7200 values in
order from 1 to 20.
If this is possible, in the rows where these values have been added, the second and third column should
changed to nan.
I realise that this is an unusual question, and that it is difficult to understand what is asked, but I hope I have been clear in expressing what i'm attempting to
acheive. Any advice would be much appreciated.
Here's one way to do it for a given element of the cell matrix:
full=zeros(7200,3)+NaN;
for i = 1:20 % for each day
starti = (i-1)*360; % find corresponding 360 indices into full array
full( starti + (1:360), 1 ) = i; % assign the day
idx = find(a(:,1)==i); % find any matching data in a for that day
full( starti + (1:length(idx)), 2:3 ) = a(idx,2:3); % copy matching data over
end
You could probably use arrayfun to make this slicker, and maybe (??) faster.
You could make this into a function and use cellfun to apply it to your cell.
PS - if you ask your question at the Matlab help forums you'll most definitely get a slicker & more efficient answer than this. Probably involving bsxfun or arrayfun or accumarray or something like that.
Update - to do this for each element in the cell array the only change is that instead of searching for i as the day number you calculate it based on how far allong the cell array you are. You'd do something like (untested):
for k = 1:length(cellarray)
for i = 1:length(cellarray{k})
starti = (i-1)*360; % ... as before
day = (k-1)*20 + i; % first cell is days 1-20, second is 21-40,...
full( starti + (1:360),1 ) = day; % <-- replace i with day
idx = find(a(:,1)==day); % <-- replace i with day
full( starti + (1:length(idx)), 2:3 ) = a(idx,2:3); % same as before
end
end
I am not sure I understood correctly what you want to do but this below works out how many measurements you are missing for each day and add at the bottom of your 'a' matrix additional lines so you do get the full 7200x3 matrix.
nbMissing = 7200-size(a,1);
a1 = nan(nbmissing,3)
l=0
for i = 1:20
nbMissing_i = 360-sum(a(:,1)=i);
a1(l+1:l+nbMissing_i,1)=i;
l = l+nb_Missing_i;
end
a_filled = [a;a1];
Could you please help me for this matter?
I have 3 matrices, P (Power), T (Temperature) and H (Humidity)
every matrix has 31 columns (days) and there are 24 rows for every column
which are the data for the March of year 2000, i.e.
for example, the matrix P has 31 columns where every column represents
a day data for Power through 24 hours and the same idea is for T and H
I tried to write a MATLAB program that accomplish my goal but
It gave me errors.
My aim is:
In the MATLAB command window, the program should ask the user the following phrase:
Please enter the day number of March, 2000 from 1 to 31:
And I know it is as follows:
Name=input (Please enter the day number of March, 2000 from 1 to 31:)
Then, when, for example, number 5 is entered, the result shown is a matrix containing the following:
1st column: The day name or it can be represented by numbers
2nd column: simple numbers from 1 to 24 representing the hours for that day
3rd column: the 24 points of P of that day extracted from the original P
(the column number 5 of the original P)
4th column: the 24 points of T of that day extracted from the original T
(the column number 5 of the original T)
5th column: the 24 points of H of that day extracted from the original H
(the column number 5 of the original H)
Any help will be highly appreciated,
Regards
Here is what you ask for:
% some sample data
P = rand(24,31);
T = rand(24,31);
H = rand(24,31);
% input day number
daynum=input('Please enter the day number of March, 2000 from 1 to 31: ');
[r, c] = size(P);
% generate output matrix
OutputMatrix = zeros(r,5);
OutputMatrix(:,1) = repmat(weekday(datenum(2000,3,daynum)),r,1);
OutputMatrix(:,2) = 1:r;
OutputMatrix(:,3) = P(:,daynum);
OutputMatrix(:,4) = T(:,daynum);
OutputMatrix(:,5) = H(:,daynum);
disp(OutputMatrix)
The matrix can be generated in a one line, but this way is clearer.
Is it always for March 2000? :) Where do you get this information from?